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LXXXIV.1 (1998)

On the average number of

unitary factors of finite abelian groups

by

J. Wu (Nancy)

1. Introduction. Let X be the semigroup of all finite abelian groups with respect to the direct product ⊗ and let E

0

be the identity of X. For G ∈ X and H ∈ X, we use (G, H) to denote the group of maximal order in X which is simultaneously a direct factor of G and H. We say that G and H are relatively prime if (G, H) = E

0

. A direct factor D of G is called unitary if D ⊗ E = G and (D, E) = E

0

. The number of unitary factors of G is denoted by t(G). In 1960, Cohen [2] proved

(1.1) X

|G|≤x

t(G) = A

1

x log x + A

2

x + O(

x log x),

where the summation is over all G in X of order |G| ≤ x and the A

j

are some effective constants. After a study on Dirichlet’s series associated with t(G), Kr¨atzel [7] found a connection between (1.1) and the following three- dimensional divisor problem:

(1.2) X

n1n2n23≤x

1 = B

1

x log x + B

2

x + B

3

x + ∆(1, 1, 2; x),

where the B

j

are some effective constants and ∆(1, 1, 2; x) is an error term. Using exponential sum techniques, he showed that ∆(1, 1, 2; x)  x

11/29

(log x)

2

, which implies

(1.3) X

|G|≤x

t(G) = A

1

x log x + A

2

x + A

3

x + ∆(x)

with ∆(x)  x

11/29

(log x)

2

. This estimate was improved to ∆(x)  x

3/8

(log x)

4

by Schmidt [11], then to ∆(x) 

ε

x

77/208+ε

by Liu [9] and to

∆(x) 

ε

x

29/80+ε

by Liu [10], where ε denotes an arbitrarily small positive number.

1991 Mathematics Subject Classification: 11L07, 11N45.

[17]

(2)

In this paper, we give a better bound.

Theorem 1. For any ε > 0, we have

∆(1, 1, 2; x) 

ε

x

47/131+ε

and ∆(x) 

ε

x

47/131+ε

.

For comparison, we have 29/80 = 0.3625 and 47/131 ≈ 0.3587. From [10], we know that in the proof of Theorem 1 the most difficult part is to estimate the exponential sums of type

(1.4) X

h∼H

a

h

X

n1∼N1

X

n2∼N2

e(xh/(n

21

n

2

)),

where |a

h

| ≤ 1, e(t) := e

2πit

and the notation h ∼ H means cH < h ≤ c

0

H with some positive unspecified constants c, c

0

. Liu [10] has treated (1.4) by combining Fouvry–Iwaniec’s method [3] and Kolesnik’s method [6].

We notice that via van der Corput’s B-process the sum (1.4) can be transformed into bilinear exponential sums of type I,

T (M, N ) := X

m∼M

X

n∈I(m)

ϕ

m

e



X m

α

n

β

M

α

N

β

 ,

where I(m) is a subinterval of [N, 2N ]. Using the classical A, B process and the well known AB theorem of Kolesnik (see Theorem 1 of [6] and Lemma 1.5 of [8]) we shall prove an estimate for T (M, N ) (see Theorem 3 below). In addition we also use an idea of Jia ([5], Lemma 13) and Liu ([8], Lemma 2.4) to investigate bilinear exponential sums of type II,

S(M, N ) := X

m∼M

X

n∼N

ϕ

m

ψ

n

e



X m

α

n

β

M

α

N

β

 .

Baker and Harman have simplified Jia–Liu’s argument to obtain a slightly more general estimate for S(M, N ) (see Theorem 2 of [1]) than those of Jia and Liu. But all such results contain some restrictions on (X, M, N ) and the number of terms is relatively large; this is not convenient in applications.

Our result (see Theorem 2 below) essentially has the same power as their estimates, but it is without restriction, more general and simpler in form.

Finally, it is worth indicating that we also need Theorem 7 of [12] and Lemma 2.3 of [13] for the proof of Theorem 1.

2. Estimates for exponential sums. We first prove two estimates for S = S(M, N ), defined as in Section 1. In the sequel, the letter ε

0

denotes a suitably small positive number (depending on α, β and α

j

at most).

Theorem 2. Let α, β ∈ R with αβ(α − 1)(β − 1) 6= 0, X > 0, M ≥ 1,

N ≥ 1, |ϕ

m

| ≤ 1, |ψ

n

| ≤ 1 and L := log(2 + XM N ). If (κ, λ) is an exponent

pair , then

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S(M, N )  {(X

2+4κ

M

8+10κ

N

9+11κ+λ

)

1/(12+16κ)

(2.1)

+ X

1/6

M

2/3

N

3/4+λ/(12+12κ)

+ (XM

3

N

4

)

1/5

+ (XM

7

N

10

)

1/11

+ M

2/3

N

11/12+λ/(12+12κ)

+ M N

1/2

+ (X

−1

M

14

N

23

)

1/22

+ X

−1/2

M N }L

2

, S(M, N )  {(XM

3

N

4

)

1/5

+ (X

4

M

10

N

11

)

1/16

+ (XM

7

N

10

)

1/11

(2.2)

+ M N

1/2

+ (X

−1

M

14

N

23

)

1/22

+ X

−1/2

M N }L

2

. P r o o f. We begin in the same way as Jia [5], Liu [8], Baker and Harman [1]. Without loss of generality, we suppose that β > 0 and L is sufficiently large. Let Q ∈ [L, N/L] be a parameter to be chosen later. By Cauchy’s inequality and a “Weyl shift” ([4], Lemma 2.5), we have

|S|

2

 (M N )

2

Q + M

3/2

N

Q

X

1≤|q1|<Q



1 − |q

1

| Q

 X

n+q1,n∼N

ψ

n+q1

ψ

n

X

m∼M

m

−1/2

e(Am

α

t),

where t = t(n, q

1

) := (n+q

1

)

β

−n

β

and A := X/(M

α

N

β

). Splitting the range of q

1

into dyadic intervals and removing 1 − q

1

/Q by partial summation, we get

(2.3) |S|

2

 (M N )

2

Q

−1

+ LM

3/2

N Q

−1

max

1≤Q1≤Q

|S(Q

1

)|, where

S(Q

1

) := X

q1∼Q1

X

n+q1,n∼N

ψ

n+q1

ψ

n

X

m∼M

m

−1/2

e(Am

α

t).

If X(M N )

−1

Q

1

≥ ε

0

, by Lemma 2.2 of [12] we can transform the inner- most sum to a sum over l and then using Lemma 2.3 of [12] with n = m we can estimate the corresponding error term. As a result, we obtain

S(Q

1

)  X

q1∼Q1

X

n+q1,n∼N

ψ

n+q1

ψ

n

X

l∈I

l

−1/2

e(e α(At)

γ

l

1−γ

) + {(XM

−1

N

−1

Q

31

)

1/2

+ M

−1/2

N Q

1

+ (X

−1

M N Q

1

)

1/2

+ (X

−2

M N

4

)

1/2

}L,

where γ := 1/(1−α), e α = |1−α|·|α|

α/(1−α)

, I := [c

1

AM

α−1

|t|, c

2

AM

α−1

|t|]

and c

j

= c

j

(α) are some constants. Exchanging the order of summation and

estimating the sum over l trivially, we find, for some l  X(M N )

−1

Q

1

, the

(4)

inequality

S(Q

1

)  (XM

−1

N

−1

Q

1

)

1/2

X X

(n,q1)∈D1(l)

ψ

n+q1

ψ

n

e(e α(At)

γ

l

1−γ

) (2.4)

+ {(XM

−1

N

−1

Q

31

)

1/2

+ M

−1/2

N Q

1

+ (X

−1

M N Q

1

)

1/2

+ (X

−2

M N

4

)

1/2

}L,

where D

1

(l) is a suitable subregion of {(n, q

1

) : n ∼ N, q

1

∼ Q

1

}. Let S

1

(Q

1

) be the double sums on the right-hand side of (2.4). Using Lem- ma 2.6 of [12] to relax the range of q

1

, we see that there exists a real number θ independent of (n, q

1

) such that

S

1

(Q

1

)  L X

n∼N

X

q1∼Q1

ψ

n+q1

e(θq

1

)e(e α(At)

γ

l

1−γ

) .

If L ≤ Q

1

≤ Q, using again Cauchy’s inequality and a “Weyl shift” with Q

2

≤ ε

0

Q

1

yields

|S

1

(Q

1

)/L|

2

 (N Q

1

)

2

Q

−12

+ N Q

1

Q

−12

X

1≤q2≤Q2

|S

2

(q

1

, q

2

)|, where

S

2

(q

1

, q

2

) := X

n∼N

X

q1+q2,q1∼Q1

ψ

n+q1+q2

ψ

n+q1

e(t

1

(n, q

1

, q

2

))

and t

1

(n, q

1

, q

2

) := e αA

γ

l

1−γ

{t(n, q

1

+ q

2

)

γ

− t(n, q

1

)

γ

}. Writing n

0

:= n + q

1

, exchanging the order of summation and using Lemma 2.6 of [12], we can deduce

S

2

(q

1

, q

2

) = X X

(n0,q1)∈D2

ψ

n0+q2

ψ

n0

e(t

1

(n

0

− q

1

, q

1

, q

2

))

 L X

n0∼N

X

q1∼Q1

e(θ

0

q

1

)e(T (n

0

, q

1

, q

2

)) ,

where T (n

0

, q

1

, q

2

) := t

1

(n

0

−q

1

, q

1

, q

2

), D

2

is a suitable subregion of {(n

0

, q

1

) : n

0

∼ N, q

1

∼ Q

1

} and θ

0

is a real number independent of (n

0

, q

1

). A final application of Cauchy’s inequality and a “Weyl shift” with Q

3

= Q

22

gives

|S

2

(q

1

, q

2

)/L|

2

 (N Q

1

)

2

Q

−13

+ N Q

1

Q

−13

X

1≤q3≤Q3

X

q1∼Q1

|S

3

(q

1

, q

2

, q

3

)|, where S

3

(q

1

, q

2

, q

3

) := P

n0∼N

e(f (n

0

)) and f (n

0

) := T (n

0

, q

1

, q

2

) − T (n

0

, q

1

+ q

3

, q

2

). It is easy to show that f (n

0

) satisfies the conditions of exponent pair and f

0

(n

0

)  XN

−2

Q

−11

q

2

q

3

(n

0

∼ N ). Hence we have

S

3

(q

1

, q

2

, q

3

)  (XN

−2

Q

−11

q

2

q

3

)

κ

N

λ

+ (XN

−2

Q

−11

q

2

q

3

)

−1

,

(5)

which implies

S

1

(Q

1

)  {(X

κ

N

3−2κ+λ

Q

4−κ1

Q

2

)

1/4

+ N Q

1

Q

−1/22

+ (X

−1

N

5

Q

51

Q

−32

)

1/4

}L

7/4

provided Q

1

≥ L, Q

2

≤ ε

0

Q

1

. By Lemma 2.4(ii) of [12] optimizing Q

2

over (0, ε

0

Q

1

] yields

S

1

(Q

1

)  {(X

κ

N

3+4κ+λ

Q

4+5κ1

)

1/(4+6κ)

+ N

3/4+λ/(4+4κ)

Q

1

+ N Q

3/41

+ (X

−2

N

10

Q

71

)

1/8

}L

7/4

provided Q

1

≥ L. In view of the term N Q

3/41

L

7/4

, this inequality holds trivially when Q

1

≤ L. Inserting the preceding estimate in (2.4) yields, for any Q

1

∈ [1, Q],

S(Q

1

)  {(X

2+4κ

M

−2−3κ

N

1+κ+λ

Q

6+8κ1

)

1/(4+6κ)

+ X

1/2

M

−1/2

N

1/4+λ/(4+4κ)

Q

3/21

+ (X

2

M

−2

N

2

Q

51

)

1/4

+ (X

2

M

−4

N

6

Q

111

)

1/8

+ (XM

−1

N

−1

Q

31

)

1/2

+ M

−1/2

N Q

1

+ (X

−1

M N Q

1

)

1/2

+ (X

−2

M N

4

)

1/2

}L

7/4

=: (E

1

+ E

2

+ . . . + E

8

)L

7/4

.

Since E

5

≤ E

3

and E

6

= (E

44

E

8

)

1/5

(M

2

Q

1

)

−1/10

, both E

5

and E

6

are superfluous. Replacing Q

1

by Q and inserting the bound obtained in (2.3), we find, for any Q ∈ [L, N/L],

S  {(X

2+4κ

M

4+6κ

N

5+7κ+λ

Q

2+2κ

)

1/(8+12κ)

(2.5)

+ X

1/4

M

1/2

N

5/8+λ/(8+8κ)

Q

1/4

+ (X

2

M

4

N

6

Q)

1/8

+ (X

2

M

8

N

14

Q

3

)

1/16

+ M N Q

−1/2

+ (X

−1

M

2

N

3

Q

−1

)

1/2

}L

11/8

,

where we have used the fact that (X

−1

M

4

N

3

Q

−1

)

1/4

can be absorbed by M N Q

−1/2

. In view of M N Q

−1/2

, the preceding estimate holds trivially when Q ∈ (0, L].

If X(M N )

−1

Q

1

≤ ε

0

, we can remove m

−1/2

by partial summation and then estimate the sum over m by Kuz’min–Landau’s inequality ([4], The- orem 2.1). Hence we see that (2.5) always holds for 0 < Q ≤ N/L. Using Lemma 2.4(ii) of [12] to optimize Q over (0, N/L] yields

S  {(X

2+4κ

M

8+10κ

N

9+11κ+λ

)

1/(12+16κ)

+ (X

M

8+10κ

N

11+13κ+λ

)

1/(12+16κ)

+ X

1/6

M

2/3

N

3/4+λ/(12+12κ)

+ M

2/3

N

11/12+λ/(12+12κ)

+ (XM

3

N

4

)

1/5

+ (XM

6

N

9

)

1/10

+ (XM

7

N

10

)

1/11

(6)

+ (X

−1

M

14

N

23

)

1/22

+ M N

1/2

+ X

−1/2

M N }L

2

=: (F

1

+ F

2

+ . . . + F

10

)L

2

. Since

F

2

= (F

46+3κ

F

5

)

1/(6+8κ)

N

−κ(1+κ−λ)/((4+4κ)(6+8κ))

and F

6

= (F

516

F

811

)

1/27

M

−2/135

, they are both superfluous. This proves (2.1).

To prove (2.2), we take Q

2

= ε

0

min{

Q

1

, (X

−1

N

2

Q

1

)

1/3

} such that

|f

0

(n

0

)| ≤ 1/2 for n

0

∼ N . Thus Kuz’min–Landau’s inequality gives S

3

(q

1

, q

2

, q

3

)  (XN

−2

Q

−11

q

2

q

3

)

−1

, from which we can deduce, as before, the following inequality:

S  {(XM

3

N

4

)

1/5

+ (XM

6

N

9

)

1/10

+ (X

4

M

10

N

11

)

1/16

+ (X

2

M

10

N

13

)

1/16

+ (XM

7

N

10

)

1/11

+ (X

−1

M

14

N

23

)

1/22

+ M N

1/2

+ X

−1/2

M N + M

1/2

N }L

2

=: (G

1

+ G

2

+ . . . + G

9

)L

2

.

It is not difficult to verify that G

2

= (G

161

G

116

)

1/27

M

−2/135

, G

4

= (G

153

G

116

)

1/26

(M

2

N

11

)

−1/416

, G

9

= (G

5

G

26

)

1/3

M

−3/22

. Thus G

2

, G

4

, G

9

are superfluous. This completes the proof.

For T = T (M, N ) defined as in Section 1, we have the following result.

Theorem 3. Let α, β ∈ R with αβ(α−1)(β−1)(α+β−1)(2α+β−2) 6= 0, X > 0, M ≥ 1, N ≥ 1, L := log(2 + XM N ), |ϕ

m

| ≤ 1 and I(m) be a subinterval of [N, 2N ]. Then

T (M, N )  {(X

5

M

10

N

8

)

1/16

+ (X

3

M

10

N

12

)

1/16

+ (XM

2

N

3

)

1/4

+ (X

3

M

14

N

18

)

1/22

+ (XM

6

N

9

)

1/10

+ (X

7

M

30

N

24

)

1/40

+ (XM

5

N

5

)

1/7

+ M N

1/2

+ X

−1

M N }L

3

.

P r o o f. If X ≤ ε

0

N , then T  X

−1

M N by Kuz’min–Landau’s inequal- ity. When X ≥ ε

0

N , using (2.3) with ψ

n

= 1, we have, for any 1 ≤ Q ≤ ε

0

N , (2.6) |T |

2

 (M N )

2

Q

−1

+ LM

3/2

N Q

−1

max

1≤Q1≤Q

|T (Q

1

)|, where

T (Q

1

) := X

q∼Q1

X

n∈I1(q)

X

m∈J(n,q)

m

−1/2

e



X m

α

t(n, q) M

α

N

β

 ,

t(n, q) := (n + q)

β

− n

β

, I

1

(q) is a subinterval of [N, 2N ] and J(n, q) a

subinterval of [M, 2M ].

(7)

If L := X(M N )

−1

Q

1

≥ ε

0

, similarly to (2.4), we can prove, for some l  L,

T (Q

1

)  (XM

−1

N

−1

Q

1

)

1/2

X X

(n,q)∈D(l)

e(f (n, q)) + {M

−1/2

N Q

1

+ (XM

−1

N

−1

Q

31

)

1/2

+ (X

−2

M N

4

)

1/2

+ (X

−1

M N Q

1

)

1/2

}L,

where f (n, q) := e α(XQ

1

/N )(l/L)

α/(α−1)

{t(n, q)/(N

β−1

Q

1

)}

1/(1−α)

and D(l) is a suitable subregion of {(n, q) : n ∼ N, q ∼ Q

1

}. It is easy to show that f (n, q) satisfies the condition of Lemma 1.5 of [8] (which is a revised form of Theorem 1 of Kolesnik [6]) with A = XN

−1

Q

1

/(N

β−1

Q

1

)

1/(1−α)

,

∆ = Q

1

/N . By this lemma with (F, X, Y ) = (XN

−1

Q

1

, N, Q

1

), we obtain the estimate

T (Q

1

)  {(X

5

M

−3

N

−2

Q

81

)

1/6

+ (X

3

M

−3

N

2

Q

81

)

1/6

(2.7)

+ (XM

−1

N Q

21

)

1/2

+ (X

3

M

−4

N

4

Q

111

)

1/8

+ (XM

−2

N

3

Q

51

)

1/4

+ (X

7

M

−5

N

−6

Q

201

)

1/10

+ (X

2

M

−2

Q

71

)

1/4

+ (X

−2

M N

4

)

1/2

+ (X

−1

M N Q

1

)

1/2

}L

4

, where we have used the fact that M

−1/2

N Q

1

+ (XM

−1

N

−1

Q

31

)

1/2

can be absorbed by (XM

−2

N

3

Q

51

)

1/4

+ (X

2

M

−2

Q

71

)

1/4

(in view of the hypothesis X ≥ ε

0

N ).

If L ≤ ε

0

, the Kuz’min–Landau inequality implies that (2.7) also holds.

Replacing Q

1

by Q and inserting into (2.6) yield

|T |

2

 {(X

5

M

6

N

4

Q

2

)

1/6

+ (X

3

M

6

N

8

Q

2

)

1/6

+ (XM

2

N

3

)

1/2

+ (X

3

M

8

N

12

Q

3

)

1/8

+ (XM

4

N

7

Q)

1/4

+ (X

7

M

10

N

4

Q

10

)

1/10

+ (X

2

M

4

N

4

Q

3

)

1/4

+ (M N )

2

Q

−1

}L

5

,

where we have eliminated two superfluous terms X

−1

M

2

N

3

Q

−1

and (X

−1

M

4

N

3

Q

−1

)

1/2

(which can be absorbed by (M N )

2

Q

−1

). Using Lem- ma 2.4(ii) of [12] to optimize Q over (0, ε

0

N ] gives the required result. This concludes the proof.

Next we shall apply Theorems 2 and 3 to treat S

I

:= X

m1∼M1

X

m2∼M2

X

m3∼M3

ψ

m2

e



X m

α11

m

α22

m

−α3 2

M

1α1

M

2α2

M

3−α2

 ,

S

II

:= X

m1∼M1

X

m2∼M2

X

m3∼M3

ϕ

m1

ψ

m2

e



X m

α11

m

α22

m

−α3 2

M

1α1

M

2α2

M

3−α2

 ,

which are general forms of (1.4). The following results will be used in the

proof of Theorem 1.

(8)

Corollary 1. Let α

j

∈ R with α

1

α

2

2

+1)(α

1

−jα

2

−j) 6= 0 (j = 1, 2), X > 0, M

j

≥ 1, |ϕ

m1

| ≤ 1, |ψ

m2

| ≤ 1 and let Y := 2 + XM

1

M

2

M

3

. If (κ, λ) is an exponent pair , then for any ε > 0,

S

II

 {(X

4+6κ

M

19+11κ+λ

M

28+10κ

M

34+6κ

)

1/(12+16κ)

+ X

1/3

M

3/4+λ/(12+12κ)

1

M

22/3

M

31/3

+ (X

3

M

18

M

26

M

34

)

1/10

+ (X

5

M

120

M

214

M

38

)

1/22

+ X

1/6

M

11/12+λ/(12+12κ)

1

M

22/3

M

31/3

+ (XM

1

M

22

)

1/2

+ (X

2

M

123

M

214

M

38

)

1/22

+ M

1

M

2

+ X

−1/2

M

2

M

3

+ X

−1

M

1

M

2

M

3

}Y

ε

. In particular , if X ≥ M

3

≥ M

1

, then

S

II

 {(X

186

M

1407

M

2350

M

3186

)

1/536

(2.8)

+ (X

164

M

1385

M

2328

M

3164

)

1/492

+ (X

3

M

18

M

26

M

34

)

1/10

+ (X

5

M

120

M

214

M

38

)

1/22

+ (XM

1

M

22

)

1/2

}Y

ε

,

S

II

 {(X

13

M

115

M

222

M

34

)

1/26

+ (X

2

M

12

M

23

M

3

)

1/4

(2.9)

+ (X

9

M

111

M

218

)

1/18

+ (XM

14

M

23

M

3

)

1/4

}Y

ε

.

P r o o f. If M

30

:= X/M

3

≤ ε

0

, the Kuz’min–Landau inequality implies S

II

 X

−1

M

1

M

2

M

3

.

Next we suppose M

30

≥ ε

0

. As before, using Lemma 2.2 of [12] to the sum over m

3

and estimating the corresponding error term by Lemma 2.3 there with n = m

1

, we obtain

S

II

 X

−1/2

M

3

S+(X

1/2

M

2

+M

1

M

2

+X

−1/2

M

2

M

3

+X

−1

M

1

M

2

M

3

) log Y, where

S := X

m1∼M1

X

m2∼M2

X

m03∼M30

e

ϕ

m1

ψ e

m2

ξ

m0

3

e

 e

α

2

X m

β11

m

β22

m

32

M

1β1

M

2β2

M

32



= X

m1∼M1

X

m02∼M20

e

ϕ

m1

ξ e

m02

e

 e

α

2

X m

β11

m

22

M

1β1

M

22

 ,

and β

j

:= α

j

/(1+α

2

) (j = 1, 2), e α

2

:= |1+α

2

|·|α

2

|

−β2

, | e ϕ

m1

| ≤ 1, | e ψ

m2

| ≤ 1,

m0

3

| ≤ 1, M

20

:= M

2

M

30

, e ξ

m0

2

:= P P

m2m03=m02

ψ e

m2

ξ

m0

3

. By Theorem 2 with (M, N ) = (M

20

, M

1

) we estimate S to get the first assertion.

In particular taking (κ, λ) = BA

2 16

,

46



=

1130

,

1630

 yields

S

II

 {(X

186

M

1407

M

2350

M

3186

)

1/536

+ (X

164

M

1385

M

2328

M

3164

)

1/492

+ (X

3

M

18

M

26

M

34

)

1/10

+ (X

5

M

120

M

214

M

38

)

1/22

+ (X

82

M

1467

M

2328

M

3164

)

1/492

+ (XM

1

M

22

)

1/2

(9)

+ (X

2

M

123

M

214

M

38

)

1/22

+ M

1

M

2

+ X

−1/2

M

2

M

3

+ X

−1

M

1

M

2

M

3

}Y

ε

=: (H

1

+ H

2

+ . . . + H

10

)Y

ε

.

Since X ≥ M

3

≥ M

1

, we have H

5

≤ H

2

, H

7

≤ H

4

, H

j

≤ H

6

(8 ≤ j ≤ 10) and thus H

5

, H

j

(7 ≤ j ≤ 10) are superfluous. This proves (2.8).

The last inequality can be proved similarly by using Theorem 7 of [12]

with (M

1

, M

2

, M

3

) = (M

1

, 1, M

20

). This completes the proof.

Corollary 2. Let α

j

∈ R with α

1

α

2

1

−1)(α

1

−2)(α

2

+1)(α

1

−α

2

−1) 6= 0, X > 0, M

j

≥ 1, |ψ

m2

| ≤ 1 and let Y := 2 + XM

1

M

2

M

3

. If M

3

≥ M

1

, then for any ε > 0 we have

S

I

 {(X

7

M

18

M

210

M

36

)

1/16

+ (X

5

M

112

M

210

M

36

)

1/16

(2.10)

+ (XM

13

M

22

M

32

)

1/4

+ (X

3

M

19

M

27

M

34

)

1/11

+ (X

2

M

19

M

26

M

34

)

1/10

+ (X

17

M

124

M

230

M

310

)

1/40

+ (X

5

M

110

M

210

M

34

)

1/14

+ (XM

1

M

22

)

1/2

+ X

−1

M

1

M

2

M

3

}Y

ε

, S

I

 {(X

15

M

111

M

222

M

34

)

1/26

+ (X

2

M

12

M

23

M

3

)

1/4

+ (X

3

M

23

M

3

)

1/4

(2.11)

+ (X

11

M

17

M

218

)

1/18

+ M

2

M

3

+ X

−1

M

1

M

2

M

3

}(log 2Y )

4

. P r o o f. As before we may suppose M

30

:= X/M

3

≥ ε

0

and prove

S

I

 X

−1/2

M

3

T (2.12)

+ (X

1/2

M

2

+ M

1

M

2

+ X

−1/2

M

2

M

3

+ X

−1

M

1

M

2

M

3

) log Y, where

T := X

m1∼M1

X

m2∼M2

X

m03∈I3

g(m

1

) e ψ

m2

ξ

m0

3

e

 e

α

2

X m

β11

m

β22

m

32

M

1β1

M

2β2

M

32

 ,

I

3

:= [c

3

(m

1

/M

1

)

α1

(m

2

/M

2

)

α2

M

30

, c

4

(m

1

/M

1

)

α1

(m

2

/M

2

)

α2

M

30

], and β

j

, e α

2

, e ψ

m2

, ξ

m0

3

are defined as before, c

j

= c

j

2

) are constants, g(m

1

) is a monomial with |g(m

1

)| ≤ 1. We define e ξ

m0

2

and M

20

in the same way as in the proof of Corollary 1. Exchanging the order of summation, we have

T = X

m02∼M20

ξ e

m0

2

X

m1∈I1(m02)

g(m

1

)e

 e

α

2

X m

β11

m

22

M

1β1

M

20β2

 ,

where I

1

(m

02

) is a subinterval of [M

1

, 2M

1

]. Removing g(m

1

) by partial

summation and estimating the double sum obtained by Theorem 3 with

(M, N ) = (M

20

, M

1

), we find

(10)

T  {(X

5

M

18

M

2010

)

1/16

+ (X

3

M

112

M

2010

)

1/16

+ (XM

13

M

202

)

1/4

+ (X

3

M

118

M

2014

)

1/22

+ (XM

19

M

206

)

1/10

+ (X

7

M

122

M

2026

)

1/36

+ (XM

15

M

205

)

1/7

+ M

11/2

M

20

}Y

ε

.

Inserting into (2.12) and noticing that the last four terms on the right- hand side of (2.12) can be absorbed by (XM

1

M

22

)

1/2

, we obtain (2.10). The inequality (2.11) is (2.7) of [13] with (M

1

, M

2

, M

3

) = (M

2

, M

1

, M

3

) and

1

, α

2

, α

3

) = (α

2

, α

1

, −α

2

). This concludes the proof.

3. Proof of Theorem 1. We shall prove only (3.1) ∆(1, 1, 2; x) 

ε

x

47/131+ε

,

since this implies ∆(x) 

ε

x

47/131+ε

by a simple convolution argument. For this we recall some standard notations. Let u := (u

1

, u

2

, u

3

) be a permuta- tion of (1, 1, 2) and let N := (N

1

, N

2

) ∈ N

2

. We write ψ(t) := {t} − 1/2 ({t}

is the fractional part of t) and define S(u, N; x) := X

1

ψ((x/(n

u11

n

u22

))

1/u3

), where the summation condition of P

1

is n

u11

n

u22+u3

≤ x, n

1

(≤)n

2

, n

1

N

1

, n

2

∼ N

2

. The notation n

1

(≤)n

2

means that n

1

= n

2

for u

1

< u

2

, and n

1

< n

2

otherwise. It is well known that for proving (3.1) it suffices to verify

S(u, N; x)  x

47/131+ε

for u = (1, 1, 2), (2, 1, 1), (1, 2, 1).

Since S(1, 1, 2, N; x)  x

5/14+ε

(see [10], p. 263), it remains to consider u = (2, 1, 1), (1, 2, 1). We shall prove the desired estimate for u = (2, 1, 1) in two cases according to the size of N

1

, which we shall formulate as two lemmas. The case of u = (1, 2, 1) can be treated similarly (more easily).

We recall that we have N

1

≤ N

2

≤ G := x/(N

12

N

2

), N

1

N

2

≤ x

1/2

when u = (2, 1, 1). This fact will be used (implicitly) many times in the proofs of Lemmas 3.1 and 3.2.

Lemma 3.1. For u = (2, 1, 1), we have

S(u, N; x) 

ε

{(x

186

N

135

)

1/536

+ (xN

12

)

1/4

+ (x

40

N

17

)

1/116

+ x

5/14

}x

ε

. In particular , if N

1

≤ x

118/655

, then S(u, N; x) 

ε

x

47/131+ε

.

P r o o f. By Lemma 2.5 of [12], we have, for any H ≥ 1, (3.2) S(u, N; x)  H

−1

N

1

N

2

+ (log x) max

1≤H0≤H

H

0−1

|S(H

0

, N)|, where

S(H

0

, N) := X

h∼H0

a

h

X

n1∼N1

X

n2∼N2

e(hx/(n

21

n

2

)), |a

h

| ≤ 1.

(11)

The inequalities (2.8) and (2.9) with (X, M

1

, M

2

, M

3

) = (GH

0

, N

1

, H

0

, N

2

) imply

S(H

0

, N)  {(G

186

N

1407

N

2186

)

1/536

+ (GN

1

H

0

)

1/2

+ χ

1

+ χ

2

}H

0

x

ε

, S(H

0

, N)  {(G

13

N

115

N

24

H

09

)

1/26

+ (G

2

N

12

N

2

H

0

)

1/4

+ (G

9

N

111

H

09

)

1/18

+ (xN

12

)

1/4

}H

0

x

ε

=: {D

1

+ D

2

+ D

3

+ (xN

12

)

1/4

}H

0

x

ε

,

with χ

1

:= (G

3

N

18

N

24

H

0−1

)

1/10

and χ

2

:= (G

5

N

120

N

28

H

0−3

)

1/22

, where we have used the fact that (G

164

N

1385

N

2164

)

1/492

≤ (G

186

N

1407

N

2186

)

1/536

(in view of N

1

≤ x

1/4

). From these, we deduce that for any H

0

≥ 1,

S(H

0

, N)  n

(G

186

N

1407

N

2186

)

1/536

+ (GN

1

H

0

)

1/2

+ (xN

12

)

1/4

+ X

1≤j≤2

X

1≤k≤3

R

j,k

o

H

0

x

ε

,

where R

j,k

:= min{χ

j

, D

k

}. Since

 

 

R

1,1

≤ (χ

451

D

131

)

1/58

= (x

40

N

17

)

1/116

, R

1,2

≤ (χ

51

D

22

)

1/7

= x

5/14

,

R

1,3

≤ (χ

51

D

3

)

1/6

= (x

36

N

111

)

1/108

,

 

 

R

2,1

≤ (χ

332

D

131

)

1/46

= (x

28

N

119

)

1/92

< x

5/14

, R

2,2

≤ (χ

112

D

62

)

1/17

= (x

11

N

14

)

1/34

< x

5/14

, R

2,3

≤ (χ

112

D

33

)

1/14

= (x

24

N

123

)

1/84

< x

5/14

, and (x

36

N

111

)

1/108

≤ (x

40

N

17

)

1/116

, we have

S(H

0

, N)  {(G

186

N

1407

N

2186

)

1/536

+ (GN

1

H

0

)

1/2

+ (xN

12

)

1/4

+ (x

40

N

17

)

1/116

+ x

5/14

}H

0

x

ε

.

Inserting into (3.2) and optimizing H by Lemma 2.4(iii) of [12] yield the desired estimate.

Lemma 3.2. For u = (2, 1, 1), we have

S(u, N; x)  {(x

7

/N

15

)

1/17

+ (x

17

/N

13

)

1/47

+ (x

11

/N

14

)

1/29

+ (x

13

/N

12

)

1/36

+ (x

5

/N

14

)

1/12

+ x

103/294

}x

ε

. In particular , if N

1

≥ x

118/655

, then S(u, N; x) 

ε

x

47/131+ε

.

P r o o f. Corollary 2 with (X, M

1

, M

2

, M

3

) = (GH

0

, N

1

, H

0

, N

2

) gives S(H

0

, N)  {L(H

0

) + (G

5

N

112

N

26

H

0−1

)

1/16

+ (GN

13

N

22

H

0−1

)

1/4

(3.3)

+ (G

3

N

19

N

24

H

0−1

)

1/11

+ (G

2

N

19

N

24

H

0−2

)

1/10

}H

0

x

ε

=: {L(H

0

) + σ

1

+ σ

2

+ σ

3

+ σ

4

}H

0

x

ε

,

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